Let $G$ be a finite group such that, for all prime number $p$, $P_p$ is a normal Sylow $p$-subgroup of $G$. Let $I$ denote the set of prime numbers dividing $|G|$ and $$K=\bigcup_{n\in\mathbb{N}}\{g\ |\ (\exists x)(x\in(\bigcup_{p\in I} P_p)^{[1,n]}\ \&\ g=x_1\ldots x_n)\}.$$
Then $G=K$. I have to prove that
$$|\prod_{p\in I} P_p|=G.$$
For $p\in I$, let $i_p:P_p\rightarrow\prod_{p\in I} P_p$ be the canonical injection. Then $\prod_{p\in I} P_p=\langle\bigcup_{p\in I} i_p(P_p)\rangle$. On the other hand, let $\phi_p:P_p\rightarrow i_p(P_p)$ be the isomorphism of $P_p$ onto $i_p(P_p)$. We can extend the family $(\phi_p)_{p\in I}$ of isomorphisms to a unique isomorphism $\phi:\cup P_p\rightarrow \cup i_p(P_p)$. This shows that $$|\bigcup P_p|=|\bigcup i_p(P_p)|.$$
Can I conclude $|\langle\bigcup_{p\in I} P_p\rangle|=|\langle\bigcup_{p\in I} i_p(P_p)\rangle|$ from this? Why or why not?
Suppose $$|G|=p_1^{k_1}p_2^{k_2}\cdots p_m^{k_m}$$ with the $p_i$ distinct primes. Then $$|P_{p_i}|=p_i^{k_i}$$ by definition of a Sylow subgroup. The order of the direct product of all of these is therefore the order of $G$.